Studies - V. Sirotkin. Program checksum.

<b>Studies</b> - V. Sirotkin. Program checksum.

(C) VA Sirotkin, g.Krasnokamensk.

 Software calculate the control

            amount.


   In some programs, direct
need is to control the accuracy and integrity of the data block 
that is loaded or saved to disk. Also

may be necessary to control the dynamic part of the program.

   This problem can be solved
checksum.

   There are different algorithms and different approaches, but 
the most effective of all is the method of cyclic redundancy 
(CRC). 

   CRC algorithm used in
IBM-programs and can detect errors up to a byte. Below is a 
listing of procedure calculates a checksum algorithm for CRC. 
(This example is taken from the magazine 'Radio' for 1992 and

given without comment).
148.
; --- Input parameters:
; HL-starting block address
; DE-end block address
; --- Imprint:
; BC-checksum block

LEN DEFB 0

START PUSH HL

        INC DE

        XOR A

        LD (LEN), A

        LD BC, 0
MET1 PUSH DE

        LD A, C

        XOR (HL)

        LD E, A

        PUSH BC

        PUSH HL

        LD BC, 00

        LD D, 8
MET2 PUSH BC

        LD A, C

        RRA

        LD A, B

        RRA

        LD B, A

        LD A, C

        RRA

        LD C, A

        POP HL

        LD A, E

        XOR L

        AND 1

        JR Z, MET3

        LD A, B

        XOR # A0

        LD B, A

        LD A, C

        XOR 1

        LD C, A
MET3 LD A, E

        RRCA

        AND # 7F

        LD E, A

        DEC D

        JR NZ, MET2

        POP HL

        POP DE

        LD A, D

        XOR C

        LD C, A

        LD A, E

        XOR B

        LD B, A

        LD A, (LEN)

        INC A

        LD (LEN), A

        INC HL

        POP DE

        LD A, H

        CP D

        JR NZ, MET1

        LD A, L

        CP E

        JR NZ, MET1

        DEC DE

        POP HL

        RET
2

   This subprogramme is moveable, ie, can operate in any email 
address unless you, of course, specify the address Tagged 'LEN'

somewhere outside of the subprogramme.

   Quite accurately tracks changes in counted byte field. Seen 
even lying next permutation of bytes. With regard to 
performance, the block length of 48 kilobytes counted for 5 
sec. 


   Arithmetic calculations.


   As you know, the ROM SOS
built a powerful program calculator that allows both simple 
arithmetic operations and actions floating point and 
trigonometric calculations. 

   But to work CALCULATOR
requires a special area - STACK
CALCULATOR, and work with
calculator is quite complicated and
tedious, especially in the programs will be overwritten by the 
entire area of RAM. 

   Sometimes, however, requires short
but effective routines that perform arithmetic operations on 
bytes. 

   Here are some
routines that allow
do without the built-in calculator. All routines can
work anywhere in RAM, ie,
relotsiruemy.


   Addition of N-byte positive
         real numbers.


   In order to add two
number, length of N bytes, it is necessary to register "B" 
Record number of bytes in the terms. Terms are entered 
originally in region with tags FIRST and SECND.

The result fits into the region with
Tagged FIRST.
148.
; Sub-adding multibyte unsigned
; Numbers

ADDN LD B, N; in the "B" - how many bytes must be

                     ; Fold

        LD DE, FIRST; address of the first term

        LD HL, SECND; address of the second term

        XOR A; carry flag will drop
SUM LD A, (DE); download the first (lowest)

                     , The number of bytes

        ADC A, (HL); add a second number

        LD (DE), A; save result

        DEC B; all bytes are composed of?

        RET Z; if all, the end

        INC HL; not, continue adding

        INC DE

        JR SUM
- -
N EQU? , How many bytes of stack?
FIRST DEFB? ; Range in length (?) Bytes

                   ; For the first day
SECND DEFB? ; Area for the second number

                   , Length (?)



     MULTIPLICATION single-byte

   positive integers.

; Multiplicand is entered in the register of 'D'
; Factor in the register of 'C'
; Result in register 'BC'

START LD B, 0, reset the high byte

                    ; Result

        LD E, 8; the number of bits in a byte
NXBIT LD A, C; factor

        RRA; the next bit in the flag 'C'

        LD C, A; return factor

        DEC E; counter diminish


        RET M; all the bits? DA-emerge


        LD A, B; byte result

        JR NC, NOADR; flags C'mnozhitelya = 0

        ADD A, D; summarize factor
NOADR RRA; move the partial sum

        LD B, A; return byte

        JR NXBIT; bit multiplication of the following items


 PROGRAM DIVISION single-byte

   Positive integers.

; Dividend is placed in register 'E'
; Divider is placed in register 'D'
; Private obtained in case 'H'
And in case 'C' produced a positive balance of
; Division

START LD HL, 08; 8 bits in a byte

       LD C, 0, reset the register balances
NEXT LD A, E; dividend shift

       RLA; left

       LD E, A; 1 bit

       LD A, C; shift in the balance

       RLA; left by 1 bit

       SUB A, D; subtract divisor

       JR NC, NOADR; balance positive?

       ADD A, D; recovery of the negative

                    ; Balance
NOADR LD C, A; tire remember

       CCF; form the bits of the private

       LD A, H; memorization

       RLA; regular

       LD H, A; numbers of private

       DEC L; decrement bit counter

       JR NZ, NEXT; cycle

       RET; output
3 routine runs in 660 cycles.



   MULTIPLICATION PROGRAM FULL

     Single-byte NUMBER OF

    Double-byte INTEGER NUMBER

     (Positive number).

2-byte multiplicand to put in the 'DE'
, 1-byte factor in the 'A'
And the result we get the registers:
, In 'A' work byte
; In 'HL' lower 16 bits of the product

START LD HL, 0, reset the register works

        LD C, 8; counter bits
NXBIT ADD HL, HL; partial sum

        RLA; shift factor

        JR NC, NOADR; bit multiplier analysis

        ADD HL, DE; summarize the multiplicand
NOADR ADC 0; account transfer

        DEC C; all 8 bits of the multiplier?

        JR NZ, NXBIT; not continue

        RET; output
;----------------------------- 2


    Binary

   Numbers in BCD.


   To display the results
number of calculations necessary
result in easy-to-output
form, ie, convert a number to
decimal code. For example, the binary number 00001111 (# 0F) is 
conveniently represented as 0001 0101

(# 15), ie convert it into
binary-decimal form. This
deals with the following routine.
148.
; BCD2B-routine transfer of 2-byte numbers.
; Binary number must be in register 'HL'.
; Result:
, In 'A' - tens of thousands;
And in 'B' - thousands and hundreds;
And in 'C' - tens and ones.
; During the procedure is called from CONV
; Sub BCD1B - translate single-byte numbers
, In which;
And in 'H' - the number to be translated, 'L' = 0
; Result: in the 'A' - level of hundreds;
And in 'B' - tens and ones

; Entrance to the procedure for converting 2-byte numbers:

BCD2B LD E, 17, counter 1 st cycle

        CALL CONV; calculate Jr. BCD byte

        LD C, A; save in 'C'

        LD E, 17; counter the 2 nd cycle

        JR PRODOL; the transition to the calculation of

; Procedure to convert 1-byte numbers

BCD1B LD E, 9; cycle for the 1-byte numbers

PRODOL CALL CONV; compute the two leading bytes

        LD B, A; save an average of bytes

        LD A, L; byte

        RET; withdraw altogether
;---- CONV XOR A; clean
SBIT DEC E; decrease the loop counter

        RET Z; the entire cycle - to leave

        ADD HL, HL; move the senior level in

                    ; Transfer

        ADC A, A

        DAA; adjust

        JR NC, SBIT; result is greater than 99?

        INC HL; yes - increased by 1

        JR SBIT; return to the calculation of
2
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